Solution of ECE 300 Test 9 F11
1.
2.
Reduce each expression to a complex number in polar form with an angle in the range −180° to +180° .
(Remember, the polar form is magnitude-angle and the magnitude cannot be negative.)
(a)
( 5∠100°)4 = 625∠400° = 625∠40°
(b)
1
1
=
= 0.0819∠ − 145°
−10 + j7 12.207∠145°
If v1 ( t ) = 9 cos ( 200π t + 30° ) and v2 ( t ) = 18sin ( 200π t − 110° ) , which sinusoid is leading and by how
many degrees? (The numerical value of the angle in degrees must lie in the range −180° to +180° .)
v1 ( t ) = 9 cos ( 200π t + 30° )
,
v2 ( t ) = 18cos ( 200π t − 200° ) = 18 cos ( 200π t + 160° )
v2 leads by 130°
3.
In the circuit below, if I L = 1∠0° find the numerical values of all the other phasor voltages and currents.
Z R = 150Ω , Z L = j120Ω , ZC = − j80Ω
VL
IR
I
s
I s = I R + I L = 1.035∠14.93° A
V
s
IL
R
L
C
V
I R = Vs / R = 0.2667∠90° A
Vs = VL + VC = j40 = 40∠90° V VL = Z L I L = j120 = 120∠90° V VC = ZC I L = − j80 = 80∠ − 90° V
C
Solution of ECE 300 Test 9 F11
1.
2.
Reduce each expression to a complex number in polar form with an angle in the range −180° to +180° .
(Remember, the polar form is magnitude-angle and the magnitude cannot be negative.)
(a)
( 4∠110°)4 = 256∠440° = 256∠80°
(b)
1
1
=
= 0.0877∠ − 127.88°
−7 + j9 11.4∠127.88°
If v1 ( t ) = 9 cos ( 200π t + 50° ) and v2 ( t ) = 18sin ( 200π t − 100° ) , which sinusoid is leading and by how
many degrees? (The numerical value of the angle in degrees must lie in the range −180° to +180° .)
v1 ( t ) = 9 cos ( 200π t + 50° )
,
v2 ( t ) = 18cos ( 200π t − 190° ) = 18 cos ( 200π t + 170° )
v2 leads by 120°
3.
In the circuit below, if I L = 1∠0° find the numerical values of all the other phasor voltages and currents.
Z R = 120Ω , Z L = j150Ω , ZC = − j90Ω
VL
IR
I
s
I s = I R + I L = 1.118∠26.57° A
V
s
IL
R
L
C
V
I R = Vs / R = 0.5∠90° A
Vs = VL + VC = j60 = 60∠90° V VL = Z L I L = j150 = 150∠90° V VC = ZC I L = − j90 = 90∠ − 90° V
C
Solution of ECE 300 Test 9 F11
1.
2.
Reduce each expression to a complex number in polar form with an angle in the range −180° to +180° .
(Remember, the polar form is magnitude-angle and the magnitude cannot be negative.)
(a)
( 3∠130°)4 = 81∠520° = 81∠160°
(b)
1
1
=
= 0.124∠ − 119.74°
−4 + j7 8.062∠119.74°
If v1 ( t ) = 9 cos ( 200π t + 20° ) and v2 ( t ) = 18sin ( 200π t − 140° ) , which sinusoid is leading and by how
many degrees? (The numerical value of the angle in degrees must lie in the range −180° to +180° .)
v1 ( t ) = 9 cos ( 200π t + 20° )
,
v2 ( t ) = 18cos ( 200π t − 230° ) = 18 cos ( 200π t + 130° )
v2 leads by 110°
3.
In the circuit below, if I L = 1∠0° find the numerical values of all the other phasor voltages and currents.
Z R = 70Ω , Z L = j60Ω , ZC = − j110Ω
VL
IR
I
s
V
s
I s = I R + I L = 1.229∠ − 35.53° A
I R = Vs / R = 0.7143∠ − 90° A
Vs = VL + VC = − j50 = 50∠ − 90° V
VL = Z L I L = j60 = 60∠90° V
IL
R
L
C
V
VC = ZC I L = − j110 = 110∠ − 90° V
C